Question

A pack of cards has one card missing. Two cards are drawn randomly and are found to be spade cards. The probability that the missing card is not a spade is

• A. \(\dfrac{39}{50}\)
• B. \(\dfrac{52}{867}\)
• C. \(\dfrac{3}{4}\)
• D. \(\dfrac{22}{425}\)

Hint:

Whenever a question asks for the probability of a cause after observing an event, use \[ \boxed{\text{Bayes' Theorem}} \] instead of direct conditional probability.

Answer 1

The Correct Option is A

Step 1: Define the events. Let \[ A=\text{the missing card is not a spade}, \] and \[ B=\text{the two drawn cards are spades}. \] We need to find \[ P(A\mid B). \]

Step 2: Find the required probabilities. Initially, \[ P(A)=\frac{39}{52}=\frac34. \] If the missing card is not a spade, then all \(13\) spades remain among \(51\) cards. Hence, \[ P(B\mid A) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{78}{1275}. \] If the missing card is a spade, then only \(12\) spades remain. Thus, \[ P(B\mid A^c) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{66}{1275}. \]

Step 3: Apply Bayes' theorem. Using \[ P(A\mid B) = \frac{P(B\mid A)P(A)} {P(B\mid A)P(A)+P(B\mid A^c)P(A^c)}, \] we get \[ P(A\mid B) = \frac{78\cdot\frac34} {78\cdot\frac34+66\cdot\frac14} = \frac{234}{300} = \frac{39}{50}. \] Therefore, \[ \boxed{\frac{39}{50}}. \] Thus, \[ \boxed{(A)} \] is the correct answer.